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MGMAT Problem (Advanced Quant)

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by [email protected] » Sun Aug 07, 2011 3:09 pm
THE ANSWER IS EEEEEEEEEEEE
IF A IS NEGATIVE THAN 1/A IS LESS THAN THE GIVEN VALUE BUT IF A IS POSITIVE THAN 1/A IS GREATER THAN THE GIVEN VALUE.
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by sumgb » Sun Aug 07, 2011 3:14 pm
THE ANSWER IS EEEEEEEEEEEE
IF A IS NEGATIVE THAN 1/A IS LESS THAN THE GIVEN VALUE BUT IF A IS POSITIVE THAN 1/A IS GREATER THAN THE GIVEN VALUE.
Hi Sandeepice,

I clearly understand you get two different answers for the question from different signs of variable a. But my real question is - can I not take square roots on both sides of second statement and conclude a is always positive.

Experts, any comments?
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by lunarpower » Sun Aug 07, 2011 11:43 pm
you can't just take square roots of even powers, unless you have previous assurance that the variables represent positive numbers.
otherwise you must use absolute values: √(n^2) = |n|.

so, the first of these statements gives |a| = |b|, which implies a = ±b.
the second gives |a| = |b^2|, or |a| = b^2, implying a = ±(b^2).

sumgb wrote:my take on this problem is ...

stmnt 1, a=b=1, true; a=b=-1 false so insuff.
hmm? how do you think it's false for a = b = -1?
if you plug a = b = -1 into the statement, then it's definitely true. (in fact, since you doing exactly the same thing to both numbers -- in this case, squaring them -- it should be clear that the statement will always be true if a and b are the same number.)
it's also true for a = 1, b = -1, and vice versa.
stmnt 2, since a^2 = b^4 this implies a = b^2 (taking square root) which means a is +ve since b^2 cant be negative.
nope.
see above.
also plug in a = b = 1;
a = b = -1;
a = 1, b = -1;
and a = -1, b = 1, and you'll quickly notice that all of them work.

the answer should be (e); this is an error in the book. it will be fixed in the next edition.
Ron has been teaching various standardized tests for 20 years.

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by gmat25 » Mon Aug 08, 2011 1:12 am
Ron, I still think Op B is the correct answer. Please correct my reasoning, here.

Given question: Is

1/a > a/(b^4 +3)

Now, as we simplify the eq'n in most of the DS question, i did the same here. Multiply the above eq'n with "a" on both sides, so u get,

1 > a^2 / (b^4 + 3)

Now, Op B says, a^2 = b^4, Moreover, both a^2 and b^4 are positive. U put any value for a^2 say any positive integer "I", u will get,

1 > I / I + 2 -----> this is always true

Now put fraction value and u will get the same result, hence i think Op B is sufficient. Please clear.
Princeton Review CAT - 710(Q-51, V-37) --> silly mistakes screwed up my VERBAL
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by lunarpower » Mon Aug 08, 2011 1:15 am
gmat25 wrote:Now, as we simplify the eq'n in most of the DS question, i did the same here. Multiply the above eq'n with "a" on both sides, so u get,

1 > a^2 / (b^4 + 3)
this step is wrong; you can't multiply an inequality by a variable of unknown sign.

if you do that, then you get ">" if a turns out to be positive, but you get "<" if a turns out to be negative.
if you want to address each of these two different inequalities according to the possibilities for a, then that approach will also work -- but it's more laborious than leaving the inequality in its current form.

to prove that your algebra is incorrect, plug a = 1, b = 1 and a = -1, b = -1 into the ORIGINAL question, and you'll see that you get one "yes" and one "no".
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by sumgb » Mon Aug 08, 2011 10:44 am
hmm? how do you think it's false for a = b = -1?
Hi Ron,
If I use a=b= -1 in the expression 1/a > a/(b^4+3); it becomes -1 > -1/4 which is a NO (not false, my mistake). whereas a=b=1 you get an answer YES.
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by lunarpower » Mon Aug 08, 2011 5:05 pm
sumgb wrote:
hmm? how do you think it's false for a = b = -1?
Hi Ron,
If I use a=b= -1 in the expression 1/a > a/(b^4+3); it becomes -1 > -1/4 which is a NO (not false, my mistake). whereas a=b=1 you get an answer YES.
ah, i thought you were saying that statement (1) itself was false if you input these numbers. ok, got it.
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by TOPGMAT » Tue Aug 09, 2011 9:45 pm
If a does not equal to zero, is 1/a > a/(b^4 +3)?

(1) a^2=b^2

(2) a^2=b^4


Can we do it this way?
if a>0 => 1 > a^2/(b^4+3) and assume a=1,b=1 => True.
if a<0 => 1 < a^2/(b^4+3) and assume a=-1,b=1 => 1 < 1/4 => False.

Hence Option A is insufficent
Similarly, Option B, can also be eliminated.
Hence E.

Thanks,
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by lunarpower » Tue Aug 09, 2011 11:43 pm
TOPGMAT wrote:If a does not equal to zero, is 1/a > a/(b^4 +3)?

(1) a^2=b^2

(2) a^2=b^4


Can we do it this way?
if a>0 => 1 > a^2/(b^4+3) and assume a=1,b=1 => True.
if a<0 => 1 < a^2/(b^4+3) and assume a=-1,b=1 => 1 < 1/4 => False.
yep -- that works.
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